Integer root of quadratic

Solution 1: We have $a = \frac{ n^2}{n-6} = n+6 + \frac{36}{n-6}$ . Hence, $n-6$ is a factor of 36. Since we are interested in positive values of $n$ , this means that $n = 2, 3, 4, 5, 7, 8, 9, 10, 12, 15, 18, 24, 42$ .

It is clear that for any of these values, $a = n+6 + \frac{36}{n-6}$ is an integer which satisfies the equation.

The sum of these values is $159$ .

Solution 2: We first ignore the condition that $n$ is positive. The equation $x^2 - ax + 6a =0$ has roots $x = \frac{ a \pm \sqrt{a^2 - 24a } } {2}$ . This is an integer if and only if $a^2 - 24a = k^2$ for some non-negative integer $k$ . (Note that $a$ and $k$ have the same parity.)

$a^2 - 24a - k^2 = 0$ has solutions $a = \frac{ 24 \pm \sqrt{24^2 + 4k^2} } { 2} = 12 \pm \sqrt{ 12^2 + k^2 }$ . Thus $12^2 + k^2 = l^2$ , or that $12^2 = l^2 - k^2 = (l-k)(l+k)$ . We can restrict our attention to non-negative integers, and since $l+k$ has the same parity as $l-k$ , we get solution pairs $(l,k) = (37,35), (20,16), (15, 9), (13,5), (12,0)$ .

This gives us pairs of $(a,k)$ as $(24,0), (0,0), (25, 5), (-1, 5), (27,9),(-3,9), (32, 16), (-8, 16), (49, 35), (-25, 35)$ .

This gives us pairs of $(a, n)$ as $(24, 12), (0, 0), (25, 25), (25, 10), (-1, 2), (-1, -3), (27, 18), (27, 9), (-3, 3),$ $(-3, -6), (32, 24), (32, 8), (-8, 4), (-8, -12), (49, 42), (49, 7), (-25, 5), (-25, -30)$ .

Hence, the sum of distinct positive $n$ is $12 + 15 + 10 + 2 + 18 + 9 + 3 + 24 + 8 + 4 + 42 + 7 + 5 = 159.$